Question

In a triangle $ABC,a=(1+\sqrt{3})cm,b=2cm$ and angle $C={60}^{\circ }$. Then the other two angles are

• A. ${45}^{\circ }$ and ${75}^{\circ }$
• B. ${30}^{\circ }$ and ${90}^{\circ }$
• C. ${105}^{\circ }$ and ${15}^{\circ }$
• D. ${100}^{\circ }$ and ${20}^{\circ }$

Answer 1

Answer: (A)

${45}^{\circ }$ and ${75}^{\circ }$

We know that, $c^2=a^2+b^2-2ab\cos C$

$={(1+\sqrt{3})}^2+2^2-2(1+\sqrt{3})\left(2\right)\cos {60}^{\circ }=6\Rightarrow c=\sqrt{6}$

From cosine rule,

$\cos B=\frac{a^2+c^2-b^2}{2ac}$ and $\cos A=\frac{b^2+c^2-a^2}{2bc}$

$\Rightarrow \cos B=\frac{1}{\sqrt{2}}$ and $\cos A=\frac{\sqrt{3}-1}{2\sqrt{2}}$

$\Rightarrow B={45}^{\circ }$ and $C={75}^{\circ }$